Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of hydrogen = 16.7 g
Mass of oxygen = 15.4 g
Limiting reactant = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 16.7 g/ 2 g/mol
Number of moles = 8.35 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 15.4 g/ 32 g/mol
Number of moles = 0.48 mol
Now we will compare the moles of both reactant with product,
H₂ : H₂O
2 : 2
8.35 : 8.35
O₂ : H₂O
1 : 2
0.48 : 2×0.48 = 0.96 mol
The number of moles of water produced by oxygen are less so it will limiting reactant.
If you mean the number of Significant Figures/Digits in 23.45, it would be 4. This is because every single non-zero digit is counted as a significant figure
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